How Do You Solve This Complex Vector Integral?

  • Context: Graduate 
  • Thread starter Thread starter hiyok
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary

Discussion Overview

The discussion revolves around solving a complex vector integral involving an exponential function and a cosine term, specifically the integral \(\int d\vec{q} e^{i\vec{q}\vec{r}} \cos(2\theta)\), where \(\theta\) is the angle of the vector \(\vec{q}\) and \(\vec{r}\) is an arbitrary vector. The integral is defined over a circular area of radius \(D\).

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • The original poster, hiyok, seeks assistance in evaluating the integral.
  • Another participant suggests using the identity for cosine, \(\cos(2\theta) = \frac{e^{2i\theta} + e^{-2i\theta}}{2}\), as a potential approach.
  • hiyok later claims to have determined that the integral amounts to zero by symmetry.
  • Another participant challenges this conclusion, referencing Wolfram Alpha, which suggests the integral is not zero.
  • There is a concern raised about the original poster's notation and the implications of the integral's domain, indicating that these factors may affect the evaluation.
  • A participant expresses regret for misunderstanding the original notation and the context of the integral.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the value of the integral, with some asserting it is zero by symmetry while others argue against this based on external computational tools.

Contextual Notes

There are unresolved issues regarding the interpretation of the notation used by the original poster and the specific conditions of the integral's domain, which may influence the evaluation.

hiyok
Messages
106
Reaction score
0
Hi,

I'm stumbled on the following integral,

\int d\vec{q} e^{i\vec{q}\vec{r}} \cos(2\theta),

where \theta denotes the angle of the vector \vec{q} and the \vec{r} is an arbitrary vector. The integral domain is a circular area of radius D.

Could anyone help me out ?

Many thanks.

hiyok
 
Physics news on Phys.org
hiyok said:
Hi,

I'm stumbled on the following integral,

\int d\vec{q} e^{i\vec{q}\vec{r}} \cos(2\theta),

where \theta denotes the angle of the vector \vec{q} and the \vec{r} is an arbitrary vector. The integral domain is a circular area of radius D.

Could anyone help me out ?

Many thanks.

hiyok

Try cos(2θ) =(exp(2iθ)+exp(-2iθ))/2
 
Thanks for your response.

I have worked it out. It amounts to zero by symmetry.
 
amind said:
> I have worked it out. It amounts to zero by symmetry.
Although , I couldn't solve but it is not zero according to wolfram alpha.
see : http://www.wolframalpha.com/input/?i=∫+(e^(i*q*r)+cos(2θ))dq
If you're going to ignore both the original poster's notation (##\theta## has a specific meaning, and it's a vector integral) and the domain of integration, you can't expect to just type the integral into Wolfram Alpha and expect to get the right result.
 
eigenperson said:
If you're going to ignore both the original poster's notation (##\theta## has a specific meaning, and it's a vector integral) and the domain of integration, you can't expect to just type the integral into Wolfram Alpha and expect to get the right result.
I'm sorry for that , I'm not great in calculus and I probably skimmed through the posts.
:(
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 11 ·
Replies
11
Views
7K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K